Question: Let $R$ be the region enclosed by the positive $x$ -axis, the positive $y$ -axis, and the curve $y=\sqrt{9-x^2}$. $y$ $x$ ${y=\sqrt{9-x^2}}$ $ R$ $ 0$ $ 3$ A solid is generated by rotating $R$ about the $x$ -axis. What is the volume of the solid? Give an exact answer in terms of $\pi$.
Explanation: Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=\sqrt{9-x^2}}$ Each slice is a cylinder. Let the thickness of each slice be $dx$ and let the radius of the base, as a function of $x$, be $r(x)$. Then, the volume of each slice is $\pi [r(x)]^2\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [r(x)]^2\,dx$ This is called the disc method. What we now need is to figure out the expression of $r(x)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=\sqrt{9-x^2}}$ $ 0$ $ 3$ $r$ The radius is equal to the distance between the curve $y=\sqrt{9-x^2}$ and the $x$ -axis. In other words, for any $x$ -value, $r(x)=\sqrt{9-x^2}}$. Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(x)}]^2 \\\\ &=\pi\left(\sqrt{9-x^2}}\right)^2 \\\\ &=\pi(9-x^2) \end{aligned}$ The leftmost endpoint of $R$ is at $x=0$ and the rightmost endpoint is at $x=3$. So the interval of integration is $[0,3]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_0^3 \left[\pi\left(9-x^2\right)\right]dx \\\\ &=\pi \int^3_0 \left(9-x^2\right)\, dx \end{aligned}$ Let's evaluate the integral. $\pi \int^3_0 \left(9-x^2\right)\, dx=18\pi$ In conclusion, the volume of the solid is $18\pi$.